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Coupled Oscillations 📂Classical Mechanics

Coupled Oscillations

Simple Coupled Oscillations

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Let’s say we have two objects, m1m_{1} and m2m_{2}, connected by two springs as shown in the above figure. Let the distance from the equilibrium point to object m1m_{1} be x1x_{1}, and to object m2m_{2} be x2x_{2}. The restoring force exerted by a spring on an object is the product of the spring constant and the stretch (or compression) of the spring, so the force exerted by spring 1 on object 1 is −k1x1-k_{1}x_{1}. Since spring 2 is compressed, it pushes object 1 to the left, so the force exerted by spring 2 on object 1 is −k2(x1−x2)-k_{2}(x_{1}-x_{2}). Therefore, the equation of motion for object 1 is as follows:

m1x1¨=−k1x1−k2(x1−x2)  ⟹  x1¨+k1+k2m1x1−k2m1x2=0 \begin{align*} && m_{1}\ddot{x_{1}} &=-k_{1}x_{1}-k_{2}(x_{1}-x_{2}) \\ \implies && \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \end{align*}

Object 2 is pulled to the left by the extension of spring 2, so the force exerted by spring 2 on object 2 is −k2(x2−x1)-k_{2}(x_{2}-x_{1}). Therefore, the equation of motion for object 2 is as follows:

m2x2¨=−k2(x2−x1)  ⟹  x2¨+k2m2x2−k2m2x1=0 \begin{align*} && m_{2}\ddot{x_{2}} &= -k_{2}(x_{2}-x_{1}) \\ \implies && \ddot{x_{2}}+\frac{k_{2}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*}

Hence, the equations of motion for the system illustrated above can be represented by the following system of coupled differential equations:

{x1¨+k1+k2m1x1−k2m1x2=0x2¨+k2m2x2−k2m2x1=0 \left\{ \begin{align*} \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \\ \ddot{x_{2}}+\frac{k_{2}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} \right.

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Coupled Oscillations with Three Springs

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Object m1m_{1} is acted upon by both spring 1 and spring 2. Calculating as before, these forces are respectively −k1x1-k_{1}x_{1} and −k2(x1−x2)-k_{2}(x_{1}-x_{2}). Therefore, the equation of motion for object 1 remains the same.

x1¨+k1+k2m1x1−k2m1x2=0 \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}=0

Object m2m_{2} is acted upon by both spring 2 and spring 3. These forces are respectively −k2(x2−x1)-k_{2}(x_{2}-x_{1}) and −k3x2-k_{3}x_{2}. Therefore, the equation of motion for object 2 is as follows:

m2x2¨=−k2(x2−x1)−k3x2  ⟹  x2¨+k2+k3m2x2−k2m2x1=0 \begin{align*} && m_{2}\ddot{x_{2}} &= -k_{2}(x_{2}-x_{1})-k_{3}x_{2} \\ \implies && \ddot{x_{2}}+\frac{k_{2}+k_{3}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*}

The overall equation of motion for the system can be represented by the following system of coupled differential equations:

{x1¨+k1+k2m1x1−k2m1x2=0x2¨+k2+k3m2x2−k2m2x1=0 \left\{ \begin{align*} \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \\ \ddot{x_{2}}+\frac{k_{2}+k_{3}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} \right.

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